Find the Only Non Repeating Element Using XOR in DSA Java - Time & Space Complexity

Time Complexity: Find the Only Non Repeating Element Using XOR

O(n)

Understanding Time Complexity

We want to know how long it takes to find the unique number in a list where every other number repeats.

How does the time needed grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int findUnique(int[] nums) {
    int result = 0;
    for (int num : nums) {
        result ^= num;
    }
    return result;
}

This code finds the only number that does not repeat by using XOR on all numbers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: XOR operation inside a single loop over the array.
• How many times: Once for each element in the array (n times).

How Execution Grows With Input

Each new number adds one XOR operation, so the work grows directly with the list size.

Input Size (n)	Approx. Operations
10	10 XOR operations
100	100 XOR operations
1000	1000 XOR operations

Pattern observation: The number of operations grows in a straight line as input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the unique number grows directly with the number of elements.

Common Mistake

[X] Wrong: "Using XOR means the time is constant no matter the list size."

[OK] Correct: Each element must be processed once, so time grows with the list size, not fixed.

Interview Connect

Understanding this helps you quickly find unique elements in data, a common task in coding challenges and real projects.

Self-Check

"What if the array was sorted? Would the time complexity change if we used a different approach?"

Practice

1. What does the XOR operation do when applied between two identical numbers?

easy

A. It returns 0

B. It returns the number itself

C. It returns 1

D. It returns -1

Solution

1. Step 1: Understand XOR property for identical numbers

   XOR of a number with itself always results in 0 because bits cancel out.

2. Step 2: Apply XOR on example

   For example, 5 XOR 5 = 0.

3. Final Answer:

   It returns 0 -> Option A

4. Quick Check:

   X XOR X = 0 [OK]

Hint: X XOR X always equals zero [OK]

Common Mistakes:

• Thinking XOR returns the number itself
• Confusing XOR with OR
• Assuming XOR returns 1 for identical numbers

2. Which of the following is the correct syntax to initialize an integer variable named result to 0 in Java?

easy

A. int result := 0;

B. int result == 0;

C. int result = 0;

D. int result = zero;

Solution

1. Step 1: Recall Java variable initialization syntax

   In Java, variables are declared with type, name, and assignment using single equals sign.

2. Step 2: Check each option

   int result = 0; uses correct syntax: int result = 0;. Others have syntax errors.

3. Final Answer:

   int result = 0; -> Option C

4. Quick Check:

   Correct Java variable initialization [OK]

Hint: Use single '=' for assignment in Java [OK]

Common Mistakes:

• Using '==' instead of '='
• Using ':=' which is not Java syntax
• Assigning undefined variable 'zero'

3. What will be the output of the following Java code?

int[] arr = {2, 3, 2, 4, 3};
int result = 0;
for (int num : arr) {
    result ^= num;
}
System.out.println(result);

medium

A. 4

B. 3

C. 2

D. 0

Solution

1. Step 1: Apply XOR operation on array elements

   Start with result=0, XOR with each element: 0^2=2, 2^3=1, 1^2=3, 3^4=7, 7^3=4.

2. Step 2: Final XOR result is the non-repeating element

   The only element that does not repeat is 4, which is the final XOR result.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Only non-repeating element = 4 [OK]

Hint: XOR all elements; result is unique number [OK]

Common Mistakes:

• Not starting XOR with 0
• Miscomputing XOR steps
• Assuming sum instead of XOR

4. Identify the error in the following Java code snippet that tries to find the only non-repeating element using XOR:

int[] arr = {1, 2, 1, 3, 2};
int result;
for (int num : arr) {
    result ^= num;
}
System.out.println(result);

medium

A. System.out.println syntax is wrong

B. Array 'arr' is not declared properly

C. For-each loop syntax is incorrect

D. Variable 'result' is not initialized before XOR operation

Solution

1. Step 1: Check variable initialization

   The variable 'result' is declared but not initialized before XOR operation, causing a compile error.

2. Step 2: Correct initialization needed

   Initialize 'result' to 0 before the loop to avoid error.

3. Final Answer:

   Variable 'result' is not initialized before XOR operation -> Option D

4. Quick Check:

   Uninitialized variable causes error [OK]

Hint: Always initialize XOR accumulator to 0 [OK]

Common Mistakes:

• Forgetting to initialize result
• Misusing loop syntax
• Incorrect print statement

5. Given an array where every element repeats twice except one element which repeats thrice, can the XOR method find the unique element? What will be the output of this code?

int[] arr = {5, 5, 5, 7, 7, 8, 8};
int result = 0;
for (int num : arr) {
    result ^= num;
}
System.out.println(result);

hard

A. 5

B. Cannot find unique element correctly

C. 0

D. 7

Solution

1. Step 1: Understand XOR property for odd/even counts

   XOR cancels even number of identical elements to 0; odd count leaves the element.

2. Step 2: Analyze given array and XOR result

   5 appears 3 times (odd) -> XOR of three identical numbers is not the number itself; XOR of 5^5=0, 0^5=5, but this pattern does not hold for triple repeats in general.

3. Step 3: Conclusion

   XOR method cannot reliably find the unique element when counts are not exactly twice except one.

4. Final Answer:

   Cannot find unique element correctly -> Option B

5. Quick Check:

   XOR fails for elements repeating thrice [OK]

Hint: XOR only works when all elements except one appear exactly twice [OK]

Common Mistakes:

• Thinking XOR finds unique element with triple repeats
• Miscomputing XOR of three identical numbers as the number itself
• Ignoring that XOR only cancels pairs